Anyone help me please?

A charge Q acts on a point charge to create an electric field. Its strength, measured a distance of 40 cm away is 100 N/C. What

KIM [24]

Answer:

$E_2=80N/C$

Explanation:

From the question we are told that:

Initial Distance $d_1=40cm=>0.4m$

Initial Electric field strength $E_1=100N/C$

Final Distance $d_2=20cm=>0.20m$

Generally the equation for Electric field is mathematically given by.

$E=\frac{kq}{d^2}$

$q=\frac{100*(0.4)^2}{K}$

Substituting q for $d=20cm$

$E_2=\frac{k}{0.2}*\frac{100*(0.4)^2}{K}$

$E_2=80N/C$

6 0

3 years ago

A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizon

notsponge [240]

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

= 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

5 0

4 years ago

1. The names of the compounds FeS, NaCl, NaOH, and Pb(CN)2 all end in the suffix A. -ite. B. -ic. C. -ide. D. -ate.

shtirl [24]

They all end with suffix "-ide"

In short, Your Answer would be Option C

Hope this helps!

5 0

3 years ago

. Prior to Remy's trip to Cleveland, his uncle tells him about this amazing barbecue restaurant there and raves about

Andrews [41]

whats the restraunt called

6 0

3 years ago

if you drop a softball from just above your knee, the kinetic energy of the ball just before it hits the ground is about 1 jule,

ipn [44]

False.

The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is
$U=mgh=(0.2 kg)(9.81 m/s^2)(0.3 m)=0.6 J$

This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.

7 0

3 years ago